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# Solve for X: 16 X − 1 = 15 X + 1 , X ≠ 0 , − 1 - CBSE Class 10 - Mathematics

ConceptRelationship Between Discriminant and Nature of Roots

#### Question

Solve for x: $\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1$

#### Solution

We have been given,$\frac{16}{x} - 1 = \frac{15}{x + 1}, x \neq 0, - 1$

Now we solve the above equation as follows,

$\frac{16 - x}{x} = \frac{15}{x + 1}$

$\Rightarrow (16 - x)(x + 1) = 15x$

$\Rightarrow 16x + 16 - x^2 - x = 15x$

$\Rightarrow 15x + 16 - x^2 - 15x = 0$

$\Rightarrow 16 - x^2 = 0$

$\Rightarrow x^2 - 16 = 0$

Now we also know that for an equation

$a x^2 + bx + c = 0$ the discriminant is given by the following equation:

$D = b^2 - 4ac$

Now, according to the equation given to us, we have, $a = 1$ ,$b = 0$ and $c = - 16$

Therefore, the discriminant is given as,

$D = (0 )^2 - 4\left( 1 \right)\left( - 16 \right)$

$= 64$

Now, the roots of an equation is given by the following equation, $x = \frac{- b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{- 0 \pm \sqrt{64}}{2\left( 1 \right)}$

$= \frac{\pm 8}{2}$

$= \pm 4$

Therefore, the value of $x = \pm 4 .$

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Solution Solve for X: 16 X − 1 = 15 X + 1 , X ≠ 0 , − 1 Concept: Relationship Between Discriminant and Nature of Roots.
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